Question

Show that the velocity of escape from the Sun at the Earth's distance from the Sun is \(\sqrt{2}\) times the speed of the Earth in its orbit, assumed to be a circle. (This is a specific case of a general result for circular orbits: \(v_{\text {esc }}=\sqrt{2} v_{\text {orb }} .\) )

Short answer

The escape velocity from the Sun at the Earth's distance from the Sun is indeed \(\sqrt{2}\) times the Earth's orbital speed.

Step-by-step solution

Establish the formula for the orbital velocity

The orbital speed of the Earth can be defined using the equation: \( v_{\text {orb }} = \sqrt{ \frac{G \cdot M_{\text {sun}}}{r}} \), where G is the gravitational constant, \( M_{\text {sun}} \) is the mass of the Sun, and r is the distance of the Earth from the Sun.

Establish the formula for the escape velocity

The escape speed of the Earth can be defined using the equation: \( v_{\text {esc }} = \sqrt{ \frac{2 \cdot G \cdot M_{\text {sun}}}{r}} \). This equation is derived from the principle of conservation of mechanical energy which implies that the kinetic energy (K.E.) of an object plus its gravitational potential energy (P.E.) is zero when the object is at an infinite distance from the Sun.

Compare the two velocities

By comparing the two formulas, it can be easily seen that: \( v_{\text {esc }} = \sqrt{2} \cdot v_{\text {orb }} \)

Key concepts

Orbital Speed

Orbital speed is the velocity at which a body must travel in order to maintain a stable orbit around another body, such as Earth orbiting the Sun. This speed is determined by the balance between the gravitational pull of the central body and the inertia of the orbiting body. In simpler terms, this is the speed needed so that the curved path of the orbiting body matches the curve of the space it is moving through, effectively allowing it to continuously 'fall' around the planet without getting any closer or moving away.

For a circular orbit, which is our consideration in the given exercise, the orbital speed is given by the formula:
\( v_{\text{orb}} = \sqrt{\frac{G \cdot M_{\text{sun}}}{r}} \)
where \( G \) is the gravitational constant, \( M_{\text{sun}} \) is the mass of the central body - in this case, the Sun, and \( r \) is the radius of the orbit, or the Earth's average distance from the Sun. The gravitational force provides the necessary centripetal force to keep the Earth in its orbit.

Gravitational Constant

The gravitational constant, denoted by \( G \), is a key quantity in Newton's law of universal gravitation. Its value is essential to calculate the force of attraction between two masses and shows up in both the formula for orbital speed and escape velocity. To put it simply, \( G \) helps us quantify the strength of gravitational attraction.

The value of \( G \) is approximately \( 6.674 \times 10^{-11} \frac{m^3}{kg \cdot s^2} \), a universal physical constant characterized by its uniformity throughout space and time. Thanks to this constant, we are able to calculate gravitational forces, orbital speeds, and escape velocities for any two masses anywhere in the universe, as long as we know their masses and the distance between them.

Conservation of Mechanical Energy

Conservation of mechanical energy is a foundational concept in physics stating that the total mechanical energy in a closed system (that is, kinetic energy plus potential energy) remains constant as long as only conservative forces (like gravity) are doing work.

In the context of escape velocity, this principle suggests that if an object were to escape the gravitational influence of another body (like the Earth from the Sun), it needs to reach a specific kinetic energy to balance out its gravitational potential energy at an infinite distance, effectively reducing its total mechanical energy to zero. The formula for escape velocity derived from this concept is:
\( v_{\text{esc}} = \sqrt{\frac{2 \cdot G \cdot M_{\text{sun}}}{r}} \)
It signifies the minimum speed that an object must have to escape the gravitational pull without further propulsion. This explains why our exercise concludes that the escape velocity is \(\sqrt{2}\) times the orbital speed; it's a direct result of conserving mechanical energy as an object moves from a stable orbit to the point of escaping the influence of the Sun.